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Now, this is your chance to use all of the AC circuit analysis techniques that

we've learned so far in the course. To figure out, in great detail, exactly

how this circuit works. So I want to take you through this

analysis. I and I really want to encourage you to

sit with a paper and pencil. Draw the circuit diagrams.

label all of the components, and write down the equations.

And try to reproduce every step that we take here.

The only way to do this, to learn this, is to do it yourself.

So, I'll, I'll take you through it. But you really need to sit down, and

figure this out on your own, at the same time.

So, here's the model of the pickup that we talked about a couple of weeks ago.

When I introduced electric guitar pickups.

So you have the, the pickup inductance from all of the coils of wire.

That wire has some resistance that is in series, then, with the inductor.

And then you have a voltage that's generated by the motion of the string in

Faraday's Law. And, but unfortunately you also have

small turn to turn capacitance, a parasitic capacitance in the pickup that

gives you a shunting capacitance. That goes across this, this whole thing.

So this was the model of the pickup and if you look back at those old videos to

see how the frequency responds, the pickup was a function of the R, L, and C

values here. Now, to analyze the circuit in more

detail, the first thing I want to do is transform this model.

Now, I'm not going to drag you through all of the detail of, of how to do this,

but you can figure out a way to represent this circuit as the circuit.

And actually, this is a more advanced topic then we're going to go into in this

course but you can use Thevenin's and Norton's Theorems.

They're really, they're just techniques for finding equivalent circuits for.

If I have a complicated circuit, I can reduce it to a simpler equa, a thevenin

or Norton equivalent circuit. Now if you want to look up those

techniques, and learn a little bit more about that.

Then, wonderful. But we just don't have time to go over

that in any detail here. But we can figure out what I want to

calculate here, is, if I look at this terminal.

And ask, what is the open circuit voltage at that terminal?

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And what is the apparent. Impedance, looking from this terminal to

ground. Then I can construct an equivalent

circuit that has the same open circuit impedance and the same impedance to

ground. I'm sorry.

It has the same open circuit voltage and the same impedance from that point to

ground. And this is the answer.

The way you figure this out, and I encourage you to do this, is assume

there's a current circulating in this loop, call it I.

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Then you can write Kirkov's Voltage Law going around this loop.

And so you have the sum of the voltage drops across the capacitor, the resistor,

the inductor, and then the voltage of the source that all has to sum up to 0.

So then you can find the current, and then the open circuit voltage is just

going to be that current times the impedence of this capacitor.

And if you work that out, you'll find the open circuit voltage comes out to be this

expression. So it's v, the original output voltage.

But then it's divided by this factor that depends upon the inductance, capacitance,

and resistance of the pickup. And it's frequency dependent.

So you write down, J omega L for this impedance, one over J omega C for that

impedance, write out the algebraic equation, solve for I, multiply by one

over J omega C and you'll find this. So, now this circuit here.

If I look into this terminal, and if I don't put a load on it.

But I just look in the, ask what's the open circuit voltage.

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Well, this voltage is going to be what I see.

because there's no current through, through these.

So there's no voltage drop. And so the open circuit voltage here is

V1, and then if I assume that the voltage, the source was zero and I look

into this terminal, this is the impedance that I see R and L in series.

That's in parallel with C. It's the same thing here.

If I replace the voltage source by a short.

In both cases, then these two circuits look the same.

So this circuit is an equivalent representation of that circuit, the

original circuit there. Now what I want to do just to simplify

this is I'll just, this RLC combination, I'm just going to.

write that as an impedance, a general impedance, Z, and that Z is the parallel

combination of the top and the bottom. While the impedance in the top up here is

R plus J omega LP. So there's the impedance of the top

branch. And then the parallel combination of two

impedances is their product. And so it's the top branch impedance

times the impedance of that capacitor. So it's the product divided by their sum.

And so, here's the first term. And there's the impedance of the

capacitor. So then I just multiply the J omega C

through the denominator here and you end up with this factor over this, okay?

So work that out. Now So now I've got the, the pickup

reduced to this relatively simple single voltage source with that voltage.

And a one impedance with this value. Now, the next step I want to take is take

my representation of the pickup. And then connect it to my volume control

circuit and figure out what's going to happen then.

Now you have to sit and stare at this a little bit, a little bit to realize that

this is really a fairly simple circuit. What you have here is you have a voltage

source, there's a pickup impedence. And then I just have another impedance to

ground. So, this thing is really just a voltage

divider. So it's the pickup impedance and then Zt

here for the tone control. That's just a series combination of Rt

and Ct. So this circuit here I can just represent

with a block zt, with these impedence. It's a series impedence of Rt plus the p

to this capacitor, which is over one over j omega ct.